Đáp án:
\[x = \frac{{34 - 15\sqrt 3 }}{8}\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge 1\)
Ta có:
\(\begin{array}{l}
5\sqrt {{x^2} - 1} + \sqrt {x + 1} + 5\sqrt {x - 1} - 3x + 1 = 0\\
\Leftrightarrow 5\sqrt {x - 1} .\sqrt {x + 1} + \sqrt {x + 1} + 5\sqrt {x - 1} + 1 - 3x = 0\\
\Leftrightarrow \sqrt {x + 1} \left( {5\sqrt {x - 1} + 1} \right) + \left( {5\sqrt {x - 1} + 1} \right) - 3x = 0\\
\Leftrightarrow \left( {\sqrt {x + 1} + 1} \right).\left( {5\sqrt {x - 1} + 1} \right) - 3x = 0\\
\Leftrightarrow \frac{{x + 1 - 1}}{{\sqrt {x + 1} - 1}}.\left( {5\sqrt {x - 1} + 1} \right) - 3x = 0\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1 \Rightarrow \sqrt {x + 1} - 1 > 0} \right)\\
\Leftrightarrow x.\left[ {\frac{{5\sqrt {x - 1} + 1}}{{\sqrt {x + 1} - 1}} - 3} \right] = 0\\
\Leftrightarrow x.\frac{{5\sqrt {x - 1} + 1 - 3\sqrt {x + 1} + 3}}{{\sqrt {x + 1} - 1}} = 0\\
\Leftrightarrow x.\frac{{5\sqrt {x - 1} - 3\sqrt {x + 1} + 4}}{{\sqrt {x + 1} - 1}} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\,\,\,\,\,\,\left( {L,\,\,x \ge 1} \right)\\
5\sqrt {x - 1} - 3\sqrt {x + 1} + 4 = 0
\end{array} \right.\\
\Leftrightarrow 5\sqrt {x - 1} + 4 = 3\sqrt {x + 1} \\
\Leftrightarrow 25\left( {x - 1} \right) + 40\sqrt {x - 1} + 16 = 9\left( {x + 1} \right)\\
\Leftrightarrow 25x - 9 + 40\sqrt {x - 1} = 9x + 9\\
\Leftrightarrow 40\sqrt {x - 1} = 18 - 16x\\
\Leftrightarrow 20\sqrt {x - 1} = 9 - 8x\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{9}{8}\\
400\left( {x - 1} \right) = 81 - 144x + 64{x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{9}{8}\\
64{x^2} - 544x + 481 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \frac{{34 \pm 15\sqrt 3 }}{8}\\
x \le \frac{9}{8}
\end{array} \right.\\
\Rightarrow x = \frac{{34 - 15\sqrt 3 }}{8}
\end{array}\)
